Answer
\begin{align*}
A & =\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
\\
C_{11} & =
\begin{vmatrix}
d
\end{vmatrix}
=d
\\
C_{12} & =-
\begin{vmatrix}
c
\end{vmatrix}
=-c
\\
C_{21} & =-
\begin{vmatrix}
b
\end{vmatrix}
=-b
\\
C_{22} & =
\begin{vmatrix}
a
\end{vmatrix}
=a
\\
\text{adj}\,A & =
\begin{bmatrix}
d & -b \\
-c & a \\
\end{bmatrix}
\\
\det A & =
\begin{vmatrix}
a & b \\
c & d \\
\end{vmatrix}
=ad-bc
\end{align*}
If $ad-bc\neq 0$, then $A$ is invertible and
\begin{align*}
A^{-1}=
\frac{\text{adj}\,A}{\det A}
=
\frac{1}{ad-bc}\begin{bmatrix}
d & -b \\
-c & a \\
\end{bmatrix}
\end{align*}
If $ad-bc=0$, then $A$ is not invertible.
Work Step by Step
\begin{align*}
A & =\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
\\
C_{11} & =
\begin{vmatrix}
d
\end{vmatrix}
=d
\\
C_{12} & =-
\begin{vmatrix}
c
\end{vmatrix}
=-c
\\
C_{21} & =-
\begin{vmatrix}
b
\end{vmatrix}
=-b
\\
C_{22} & =
\begin{vmatrix}
a
\end{vmatrix}
=a
\\
\text{adj}\,A & =
\begin{bmatrix}
d & -b \\
-c & a \\
\end{bmatrix}
\\
\det A & =
\begin{vmatrix}
a & b \\
c & d \\
\end{vmatrix}
=ad-bc
\end{align*}
If $ad-bc\neq 0$, then $A$ is invertible and
\begin{align*}
A^{-1}=
\frac{\text{adj}\,A}{\det A}
=
\frac{1}{ad-bc}\begin{bmatrix}
d & -b \\
-c & a \\
\end{bmatrix}
\end{align*}
If $ad-bc=0$, then $A$ is not invertible.